1
00:00:00 --> 00:00:06

2
00:00:40 --> 00:00:46
This is also written in the
form, it's the k that's on the

3
00:00:45 --> 00:00:51
right hand side.
Actually, I found that source

4
00:00:49 --> 00:00:55
is of considerable difficulty.
And, in general,

5
00:00:53 --> 00:00:59
it is.
For these, the temperature

6
00:00:56 --> 00:01:02
concentration model,
it's natural to have the k on

7
00:01:01 --> 00:01:07
the right-hand side,
and to separate out the (q)e as

8
00:01:05 --> 00:01:11
part of it.
Another model for which that's

9
00:01:10 --> 00:01:16
true is mixing,
as I think I will show you on

10
00:01:14 --> 00:01:20
Monday.
On the other hand,

11
00:01:16 --> 00:01:22
there are some common
first-order models for which

12
00:01:21 --> 00:01:27
it's not a natural way to
separate things out.

13
00:01:25 --> 00:01:31
Examples would be the RC
circuit, radioactive decay,

14
00:01:29 --> 00:01:35
stuff like that.
So, this is not a universal

15
00:01:34 --> 00:01:40
utility.
But I thought that that form of

16
00:01:38 --> 00:01:44
writing it was a sufficient
utility to make a special case,

17
00:01:43 --> 00:01:49
and I emphasize it very heavily
in the nodes.

18
00:01:47 --> 00:01:53
Let's look at the equation.
And, this form will be good

19
00:01:52 --> 00:01:58
enough, the y prime.
When you solve it,

20
00:01:56 --> 00:02:02
let me remind you how the
solutions look,

21
00:02:00 --> 00:02:06
because that explains the
terminology.

22
00:02:05 --> 00:02:11
The solution looks like,
after you have done the

23
00:02:08 --> 00:02:14
integrating factor and
multiplied through,

24
00:02:11 --> 00:02:17
and integrated both sides,
in short, what you're supposed

25
00:02:16 --> 00:02:22
to do, the solution looks like y
equals, there's the term e to

26
00:02:20 --> 00:02:26
the negative k out
front times an integral which

27
00:02:25 --> 00:02:31
you can either make definite or
indefinite, according to your

28
00:02:30 --> 00:02:36
preference.
q of t times e to the kt inside

29
00:02:34 --> 00:02:40
dt,
it will help you to remember

30
00:02:37 --> 00:02:43
the opposite signs if you think
that when q is a constant,

31
00:02:41 --> 00:02:47
one, for example,
you want these two guys to

32
00:02:44 --> 00:02:50
cancel out and produce a
constant solution.

33
00:02:47 --> 00:02:53
That's a good way to remember
that the signs have to be

34
00:02:50 --> 00:02:56
opposite.
But, I don't encourage you to

35
00:02:53 --> 00:02:59
remember the formula at all.
It's just a convenient thing

36
00:02:56 --> 00:03:02
for me to be able to use right
now.

37
00:03:00 --> 00:03:06
And then, there's the other
term, which comes by putting up

38
00:03:04 --> 00:03:10
the arbitrary constant
explicitly, c e to the negative

39
00:03:09 --> 00:03:15
kt.
So, you could either write it

40
00:03:12 --> 00:03:18
this way, where this is somewhat
vague, or you could make it

41
00:03:17 --> 00:03:23
definite by putting a zero here
and a t there,

42
00:03:21 --> 00:03:27
and change the dummy variable
inside according to the way the

43
00:03:26 --> 00:03:32
notes tell you to do it.
Now, when you do this,

44
00:03:31 --> 00:03:37
and if k is positive,
that's absolutely essential,

45
00:03:35 --> 00:03:41
only when that is so,
then this term,

46
00:03:39 --> 00:03:45
as I told you a week or so ago,
this term goes to zero because

47
00:03:45 --> 00:03:51
k is positive as t goes to
infinity.

48
00:03:48 --> 00:03:54
So, this goes to zero as t
goes, and it doesn't matter what

49
00:03:53 --> 00:03:59
c is, as t goes to infinity.
This term stays some sort of

50
00:03:59 --> 00:04:05
function.
And so, this term is called the

51
00:04:03 --> 00:04:09
steady-state or long-term
solution, or it's called both,

52
00:04:08 --> 00:04:14
a long-term solution.
And this, which disappears,

53
00:04:14 --> 00:04:20
gets smaller and smaller as
time goes on,

54
00:04:17 --> 00:04:23
is therefore called the
transient because it disappears

55
00:04:21 --> 00:04:27
at the time increases to
infinity.

56
00:04:24 --> 00:04:30
So, this part uses the initial
condition, uses the initial

57
00:04:28 --> 00:04:34
value.
Let's call it y of zero,

58
00:04:31 --> 00:04:37
assuming that you
started the initial value,

59
00:04:35 --> 00:04:41
t, when t was equal to zero,
which is a common thing to do,

60
00:04:40 --> 00:04:46
although of course not
necessary.

61
00:04:44 --> 00:04:50
The starting value appears in
this term.

62
00:04:47 --> 00:04:53
This one is just some function.
Now, the general picture or the

63
00:04:53 --> 00:04:59
way that looks is,
the steady-state solution will

64
00:04:57 --> 00:05:03
be some solution like,
I don't know,

65
00:05:01 --> 00:05:07
like that, let's say.
So, that's a steady-state

66
00:05:05 --> 00:05:11
solution, the SSS.
Well, what do the other guys

67
00:05:09 --> 00:05:15
look like?
Well, the steady-state solution

68
00:05:12 --> 00:05:18
has this starting point.
Other solutions can have any of

69
00:05:16 --> 00:05:22
these other starting points.
So, in the beginning,

70
00:05:20 --> 00:05:26
they won't look like the
steady-state solution.

71
00:05:23 --> 00:05:29
But, we know that as time goes
on, they must approach it

72
00:05:27 --> 00:05:33
because this term represents the
difference between the solution

73
00:05:31 --> 00:05:37
and the steady-state solution.
So, this term is going to zero.

74
00:05:37 --> 00:05:43
And therefore,
whatever these guys do to start

75
00:05:41 --> 00:05:47
out with, after a while they
must follow the steady-state

76
00:05:45 --> 00:05:51
solution more and more closely.
They must, in short,

77
00:05:49 --> 00:05:55
be asymptotic to it.
So, the solutions to any

78
00:05:53 --> 00:05:59
equation of that form will look
like this.

79
00:05:56 --> 00:06:02
Up here, maybe it started at
127.

80
00:05:58 --> 00:06:04
That's okay.
After a while,

81
00:06:00 --> 00:06:06
it's going to start approaching
that green curve.

82
00:06:06 --> 00:06:12
Of course, they won't cross
each other.

83
00:06:09 --> 00:06:15
That's the rock star,
and these are the groupies

84
00:06:13 --> 00:06:19
trying to get close to it.
Now, but something follows from

85
00:06:18 --> 00:06:24
that picture.
Which is the steady-state

86
00:06:22 --> 00:06:28
solution?
What, in short,

87
00:06:24 --> 00:06:30
is so special about this green
curve?

88
00:06:27 --> 00:06:33
All these other white solution
curves have that same property,

89
00:06:33 --> 00:06:39
the same property that all the
other white curves and the green

90
00:06:38 --> 00:06:44
curve, too, are trying to get
close to them.

91
00:06:44 --> 00:06:50
In other words,
there is nothing special about

92
00:06:47 --> 00:06:53
the green curve.
It's just that they all want to

93
00:06:51 --> 00:06:57
get close to each other.
And therefore,

94
00:06:54 --> 00:07:00
though you can write a formula
like this, there isn't one

95
00:06:58 --> 00:07:04
steady-state solution.
There are many.

96
00:07:01 --> 00:07:07
Now, this produces vagueness.
You talk about the steady-state

97
00:07:07 --> 00:07:13
solution; which one are you
talking about?

98
00:07:09 --> 00:07:15
I have no answer to that;
the usual answer is whichever

99
00:07:13 --> 00:07:19
one looks simplest.
Normally, the one that will

100
00:07:16 --> 00:07:22
look simplest is the one where c
is zero.

101
00:07:19 --> 00:07:25
But, if this is a peculiar
function, it might be that for

102
00:07:23 --> 00:07:29
some other value of c,
you get an even simpler

103
00:07:26 --> 00:07:32
expression.
So, the steady-state solution:

104
00:07:29 --> 00:07:35
about the best I can see,
either you integrate that,

105
00:07:32 --> 00:07:38
don't use an arbitrary
constant, and use what you get,

106
00:07:36 --> 00:07:42
or pick the simplest.
Pick the value of c,

107
00:07:41 --> 00:07:47
which gives you the simplest
answer.

108
00:07:46 --> 00:07:52
Pick the simplest function,
and that's what usually called

109
00:07:53 --> 00:07:59
the steady-state solution.
Now, from that point of view,

110
00:08:00 --> 00:08:06
what I'm calling the input in
this input response point of

111
00:08:05 --> 00:08:11
view, which we are going to be
using, by the way,

112
00:08:08 --> 00:08:14
constantly, well,
pretty much all term long,

113
00:08:12 --> 00:08:18
but certainly for the next
month or so, I'm constantly

114
00:08:16 --> 00:08:22
going to be coming back to it.
The input is the q of t.

115
00:08:21 --> 00:08:27
In other words,

116
00:08:23 --> 00:08:29
it seems rather peculiar.
But the input is the right-hand

117
00:08:27 --> 00:08:33
side of the equation of the
differential equation.

118
00:08:31 --> 00:08:37
And the reason is because I'm
always thinking of the

119
00:08:35 --> 00:08:41
temperature model.
The external water bath at

120
00:08:41 --> 00:08:47
temperature T external,
the internal thing here,

121
00:08:44 --> 00:08:50
the problem is,
given this function,

122
00:08:47 --> 00:08:53
the external water bath
temperature is driving,

123
00:08:51 --> 00:08:57
so to speak,
the temperature of the inside.

124
00:08:54 --> 00:09:00
And therefore,
the input is the temperature of

125
00:08:57 --> 00:09:03
the water bath.
I don't like the word output,

126
00:09:01 --> 00:09:07
although it would be the
natural thing because this

127
00:09:05 --> 00:09:11
temperature doesn't look like an
output.

128
00:09:07 --> 00:09:13
Anyone might be willing to say,
yeah, you are inputting the

129
00:09:11 --> 00:09:17
value of the temperature here.
This, it's more likely,

130
00:09:15 --> 00:09:21
the normal term is response.
This thing, this plus the water

131
00:09:19 --> 00:09:25
bath, is a little system.
And the response of the system,

132
00:09:22 --> 00:09:28
i.e.
the change in the internal

133
00:09:24 --> 00:09:30
temperature is governed by the
driving external temperature.

134
00:09:28 --> 00:09:34
So, the input is q of t,
and the response of

135
00:09:32 --> 00:09:38
the system is the solution to
the differential equation.

136
00:09:37 --> 00:09:43

137
00:09:45 --> 00:09:51
Now, if the thing is special,
as it's going to be for most of

138
00:09:49 --> 00:09:55
this period, it has that special
form, then I'm going to,

139
00:09:54 --> 00:10:00
I really want to call q sub e
the input.

140
00:09:58 --> 00:10:04
I want to call q sub e the
input, and there is no standard

141
00:10:02 --> 00:10:08
way of doing that,
although there's a most common

142
00:10:06 --> 00:10:12
way.
So, I'm just calling it the

143
00:10:10 --> 00:10:16
physical input,
in other words,

144
00:10:12 --> 00:10:18
the temperature input,
or the concentration input.

145
00:10:16 --> 00:10:22
And, that will be my (q)e of t,
and by the

146
00:10:21 --> 00:10:27
subscript e, you will understand
that I'm writing it in that form

147
00:10:26 --> 00:10:32
and thinking of this model,
or concentration model,

148
00:10:30 --> 00:10:36
or mixing model as I will show
you on Monday.

149
00:10:35 --> 00:10:41
By the way, this is often
handled, I mean,

150
00:10:37 --> 00:10:43
how would you handle this to
get rid of a k?

151
00:10:41 --> 00:10:47
Well, divide through by k.
So, this equation is often,

152
00:10:45 --> 00:10:51
in the literature,
written this way:

153
00:10:47 --> 00:10:53
one over k times y prime plus y
is equal to, well,

154
00:10:51 --> 00:10:57
now they call it q of t,
not (q)e of t because they've

155
00:10:55 --> 00:11:01
gotten rid of this funny factor.
But

156
00:10:59 --> 00:11:05
I will continue to call it (q)e.
So, in other words,

157
00:11:04 --> 00:11:10
and this part this is just,
frankly, called the input.

158
00:11:09 --> 00:11:15
It doesn't say physical or
anything.

159
00:11:11 --> 00:11:17
And, this is the solution,
it's then the response,

160
00:11:16 --> 00:11:22
and this funny coefficient of y
prime,

161
00:11:19 --> 00:11:25
that's not in standard linear
form, is it, anymore?

162
00:11:23 --> 00:11:29
But, it's a standard form if
you want to do this input

163
00:11:28 --> 00:11:34
response analysis.
So, this is also a way of

164
00:11:31 --> 00:11:37
writing the equation.
I'm not going to use it because

165
00:11:38 --> 00:11:44
how many standard forms could
this poor little course absorb?

166
00:11:43 --> 00:11:49
I'll stick to that one.
Okay, you have,

167
00:11:47 --> 00:11:53
then, the superposition
principle, which I don't think

168
00:11:52 --> 00:11:58
I'm going to-- the solution,
which solution?

169
00:11:57 --> 00:12:03
Well, normally it means any
solution, or in other words,

170
00:12:02 --> 00:12:08
the steady-state solution.
Now, notice that terminology

171
00:12:07 --> 00:12:13
only makes sense if k is
positive.

172
00:12:10 --> 00:12:16
And, in fact,
there is nothing like the

173
00:12:13 --> 00:12:19
picture, the picture doesn't
look at all like this if k is

174
00:12:17 --> 00:12:23
negative, and therefore,
the terms would steady state,

175
00:12:20 --> 00:12:26
transient would be totally
inappropriate if k were

176
00:12:24 --> 00:12:30
negative.
So, this assumes definitely

177
00:12:26 --> 00:12:32
that k has to be greater than
zero.

178
00:12:30 --> 00:12:36
Otherwise, no.
So, I'll call this the physical

179
00:12:33 --> 00:12:39
input.
And then, you have the

180
00:12:35 --> 00:12:41
superposition principle,
which I really can't improve

181
00:12:40 --> 00:12:46
upon what's written in the
notes, this superposition of

182
00:12:44 --> 00:12:50
inputs.
Whether they are physical

183
00:12:47 --> 00:12:53
inputs or nonphysical inputs,
if the input q of t produces

184
00:12:51 --> 00:12:57
the response, y of t,

185
00:12:54 --> 00:13:00
and q two of t produces the
response, y two of t,

186
00:12:59 --> 00:13:05

187
00:13:02 --> 00:13:08
-- then a simple calculation
with the differential equation

188
00:13:07 --> 00:13:13
shows you that by,
so to speak,

189
00:13:10 --> 00:13:16
adding, that the sum of these
two, I stated it very generally

190
00:13:15 --> 00:13:21
in the notes but it corresponds,
we will have as the response

191
00:13:21 --> 00:13:27
y1, the steady-state response y1
plus y2,

192
00:13:25 --> 00:13:31
and a constant times y1.
That's an expression,

193
00:13:30 --> 00:13:36
essentially,
of the linear,

194
00:13:32 --> 00:13:38
it uses the fact that the
special form of the equation,

195
00:13:35 --> 00:13:41
and we will have a very
efficient and elegant way of

196
00:13:38 --> 00:13:44
seeing this when we study higher
order equations.

197
00:13:41 --> 00:13:47
For now, I will just,
the little calculation that's

198
00:13:45 --> 00:13:51
done in the notes will suffice
for first-order equations.

199
00:13:48 --> 00:13:54
If you don't have a complicated
equation, there's no point in

200
00:13:52 --> 00:13:58
making a fuss over proofs using
it.

201
00:13:54 --> 00:14:00
But essentially,
it uses the fact that the

202
00:13:57 --> 00:14:03
equation is linear.
Or, that's bad,

203
00:14:01 --> 00:14:07
so linearity of the ODE.
In other words,

204
00:14:04 --> 00:14:10
it's a consequence of the fact
that the equation looks the way

205
00:14:09 --> 00:14:15
it does.
And, something like this would

206
00:14:12 --> 00:14:18
not, in any sense,
be true if the equation,

207
00:14:15 --> 00:14:21
for example,
had here a y squared

208
00:14:18 --> 00:14:24
instead of t.
Everything I'm saying this

209
00:14:21 --> 00:14:27
period would be total nonsense
and totally inapplicable.

210
00:14:27 --> 00:14:33
Now, today, what I wanted to
discuss was, what's in the notes

211
00:14:32 --> 00:14:38
that I gave you today,
which is, what happens when the

212
00:14:36 --> 00:14:42
physical input is trigonometric?
For certain reasons,

213
00:14:41 --> 00:14:47
that's the most important case
there is.

214
00:14:44 --> 00:14:50
It's because of the existence
of what are called Fourier

215
00:14:49 --> 00:14:55
series, and there are a couple
of words about them.

216
00:14:53 --> 00:14:59
That's something we will be
studying in about three weeks or

217
00:14:58 --> 00:15:04
so.
What's going on,

218
00:15:01 --> 00:15:07
roughly, is that,
so I'm going to take the

219
00:15:06 --> 00:15:12
equation in the form y prime
plus ky equals k times 

220
00:15:12 --> 00:15:18
(q)e of t,
and the input that I'm

221
00:15:17 --> 00:15:23
interested in is when this is a
simple one that you use on the

222
00:15:23 --> 00:15:29
visual that you did about two
points worth of work for handing

223
00:15:30 --> 00:15:36
in today, cosine omega t.

224
00:15:36 --> 00:15:42
So, if you like,
k here.

225
00:15:37 --> 00:15:43
So, the (q)e is cosine omega t.
That was the physical input.

226
00:15:41 --> 00:15:47
And, omega, as you know,
is, you have to be careful when

227
00:15:44 --> 00:15:50
you use the word frequency.
I assume you got this from

228
00:15:48 --> 00:15:54
physics class all last semester.
But anyway, just to remind you,

229
00:15:52 --> 00:15:58
there's a whole yoga of five or
six terms that go whenever

230
00:15:56 --> 00:16:02
you're talking about
trigonometric functions.

231
00:16:00 --> 00:16:06
Instead of giving a long
explanation, the end of the

232
00:16:03 --> 00:16:09
second page of the notes just
gives you a reference list of

233
00:16:08 --> 00:16:14
what you are expected to know
for 18.03 and physics as well,

234
00:16:12 --> 00:16:18
with a brief one or two line
description of what each of

235
00:16:17 --> 00:16:23
those means.
So, think of it as something to

236
00:16:20 --> 00:16:26
refer back to if you have
forgotten.

237
00:16:23 --> 00:16:29
But, omega is what's called the
angular frequency or the

238
00:16:27 --> 00:16:33
circular frequency.
It's somewhat misleading to

239
00:16:31 --> 00:16:37
call it the frequency,
although I probably will.

240
00:16:36 --> 00:16:42
It's the angular frequency.
It's, in other words,

241
00:16:39 --> 00:16:45
it's the number of complete
oscillations.

242
00:16:42 --> 00:16:48
This cosine omega t
is going up and down right?

243
00:16:47 --> 00:16:53
So, a complete oscillation as
it goes down and then returns to

244
00:16:51 --> 00:16:57
where it started.
That's a complete oscillation.

245
00:16:55 --> 00:17:01
This is only half an
oscillation because you didn't

246
00:16:58 --> 00:17:04
give it a chance to get back.
Okay, so the number of complete

247
00:17:03 --> 00:17:09
oscillations in how much time,
well, in two pi,

248
00:17:06 --> 00:17:12
in the distance,
two pi on the t-axis in the

249
00:17:09 --> 00:17:15
interval of length two pi
because, for example,

250
00:17:13 --> 00:17:19
if omega is one,
cosine t takes two

251
00:17:16 --> 00:17:22
pi to repeat itself,
right?

252
00:17:20 --> 00:17:26
If omega were two,
it would repeat itself.

253
00:17:23 --> 00:17:29
It would make two complete
oscillations in the interval,

254
00:17:28 --> 00:17:34
two pi.
So, it's what happens to the

255
00:17:31 --> 00:17:37
interval, two pi,
not what happens in the time

256
00:17:35 --> 00:17:41
interval, one,
which is the natural meaning of

257
00:17:39 --> 00:17:45
the word frequency.
There's always this factor of

258
00:17:43 --> 00:17:49
two pi that floats around to
make all of your formulas and

259
00:17:48 --> 00:17:54
solutions incorrect.
Okay, now, so,

260
00:17:51 --> 00:17:57
what I'm out to do is,
the problem is for the physical

261
00:17:55 --> 00:18:01
input, (q)e cosine omega t,

262
00:17:59 --> 00:18:05
find the response.
In other words,

263
00:18:02 --> 00:18:08
solve the differential
equation.

264
00:18:07 --> 00:18:13
In short, for the visual that
you looked at,

265
00:18:11 --> 00:18:17
I think I've forgot the colors
now.

266
00:18:14 --> 00:18:20
The input was in green,
maybe, but I do remember that

267
00:18:19 --> 00:18:25
the response was in yellow.
I think I remember that.

268
00:18:24 --> 00:18:30
So, find the response,
yellow, and the input was,

269
00:18:28 --> 00:18:34
what color was it,
green?

270
00:18:30 --> 00:18:36
Blue, blue.
Light blue.

271
00:18:34 --> 00:18:40
Okay, so we've got to solve the
differential equation.

272
00:18:39 --> 00:18:45
Now, it's a question of how I'm
going to solve the differential

273
00:18:46 --> 00:18:52
equation.
I'm going to use complex

274
00:18:49 --> 00:18:55
numbers throughout,
A because that's the way it's

275
00:18:54 --> 00:19:00
usually done.
B, to give you practice using

276
00:18:59 --> 00:19:05
complex numbers,
and I don't think I need any

277
00:19:04 --> 00:19:10
other reasons.
So, I'm going to use complex

278
00:19:09 --> 00:19:15
numbers.
I'm going to complexify.

279
00:19:13 --> 00:19:19
To use complex numbers,
what you do is complexification

280
00:19:18 --> 00:19:24
of the problem.
So, I'm going to complexify the

281
00:19:23 --> 00:19:29
problem, turn it into the domain
of complex numbers.

282
00:19:29 --> 00:19:35
So, take the differential
equation, turn it into a

283
00:19:33 --> 00:19:39
differential equation involving
complex numbers,

284
00:19:37 --> 00:19:43
solve that, and then go back to
the real domain to get the

285
00:19:42 --> 00:19:48
answer, since it's easier to
integrate exponentials.

286
00:19:46 --> 00:19:52
And therefore,
try to introduce,

287
00:19:49 --> 00:19:55
try to change the trigonometric
functions into complex

288
00:19:53 --> 00:19:59
exponentials,
simply because the work will be

289
00:19:57 --> 00:20:03
easier to do.
All right, so let's do it.

290
00:20:02 --> 00:20:08
To change this differential
equation, remember,

291
00:20:05 --> 00:20:11
I've got cosine omega t here.

292
00:20:09 --> 00:20:15
I'm going to use the fact that
e to the i omega t,

293
00:20:14 --> 00:20:20
Euler's formula,
that the real part of it is

294
00:20:17 --> 00:20:23
cosine omega t.
So, I'm going to view this as

295
00:20:22 --> 00:20:28
the real part of this complex
function.

296
00:20:25 --> 00:20:31
But, I will throw at the
imaginary part,

297
00:20:28 --> 00:20:34
too, since at one point we will
need it.

298
00:20:31 --> 00:20:37
Now, what is the equation,
then, that it's going to turn

299
00:20:36 --> 00:20:42
into?
The complexified equation is

300
00:20:41 --> 00:20:47
going to be y prime plus ky
equals, and now,

301
00:20:46 --> 00:20:52
instead of the right hand side,
k times cosine omega t,

302
00:20:53 --> 00:20:59
I'll use the whole complex
exponential, e i omega t.

303
00:21:00 --> 00:21:06
Now, I have a problem because

304
00:21:06 --> 00:21:12
y, here, in this equation,
y means the real function which

305
00:21:09 --> 00:21:15
solves that problem.
I therefore cannot continue to

306
00:21:13 --> 00:21:19
call this y because I want y to
be a real function.

307
00:21:16 --> 00:21:22
I have to change its name.
Since this is complex function

308
00:21:20 --> 00:21:26
on the right-hand side,
I will have to expect a complex

309
00:21:24 --> 00:21:30
solution to the differential
equation.

310
00:21:28 --> 00:21:34
I'm going to call that complex
solution y tilda.

311
00:21:32 --> 00:21:38
Now, that's what I would also
use as the designation for the

312
00:21:38 --> 00:21:44
variable.
So, y tilda is the complex

313
00:21:42 --> 00:21:48
solution.
And, it's going to have the

314
00:21:46 --> 00:21:52
form y1 plus i times y2.

315
00:21:49 --> 00:21:55
It's going to be the complex
solution.

316
00:21:53 --> 00:21:59
And now, what I say is,
so, solve it.

317
00:21:57 --> 00:22:03
Find this complex solution.
So, find the program is to find

318
00:22:03 --> 00:22:09
y tilde, --
-- that's the complex solution.

319
00:22:08 --> 00:22:14
And then I say,
all you have to do is take the

320
00:22:12 --> 00:22:18
real part of that,
and that will answer the

321
00:22:16 --> 00:22:22
original problem.
Then, y1, that's the real part

322
00:22:20 --> 00:22:26
of it, right?
It's a function,

323
00:22:23 --> 00:22:29
you know, like this is cosine
plus sine, as it was over here,

324
00:22:28 --> 00:22:34
it will naturally be something
different.

325
00:22:31 --> 00:22:37
It will be something different,
but that part of it,

326
00:22:36 --> 00:22:42
the real part will solve the
original problem,

327
00:22:40 --> 00:22:46
the original,
real, ODE.

328
00:22:44 --> 00:22:50
Now, you will say,
you expect us to believe that?

329
00:22:47 --> 00:22:53
Well, yes, in fact.
I think we've got a lot to do,

330
00:22:51 --> 00:22:57
so since the argument for this
is given in the nodes,

331
00:22:54 --> 00:23:00
so, read this in the notes.
It only takes a line or two of

332
00:22:58 --> 00:23:04
standard work with
differentiation.

333
00:23:02 --> 00:23:08
So, read in the notes the
argument for that,

334
00:23:05 --> 00:23:11
why that's so.
It just amounts to separating

335
00:23:09 --> 00:23:15
real and imaginary parts.
Okay, so let's,

336
00:23:13 --> 00:23:19
now, solve this.
Since that's our program,

337
00:23:17 --> 00:23:23
all we have to find is the
solution.

338
00:23:20 --> 00:23:26
Well, just use integrating
factors and just do it.

339
00:23:25 --> 00:23:31
So, the integrating factor will
be, what, e to the,

340
00:23:29 --> 00:23:35
I don't want to use that
formula.

341
00:23:34 --> 00:23:40
So, the integrating factor will
be e to the kt is the

342
00:23:38 --> 00:23:44
integrating factor.
If I multiply through both

343
00:23:42 --> 00:23:48
sides by the integrating factor,
then the left-hand side will

344
00:23:46 --> 00:23:52
become y e to the kt,
the way it always does,

345
00:23:50 --> 00:23:56
prime, Y tilde, sorry,

346
00:23:53 --> 00:23:59
and the right-hand side will
be, now I'm going to start

347
00:23:57 --> 00:24:03
combining exponentials.
It will be k times e to the

348
00:24:03 --> 00:24:09
power i times omega t plus k.

349
00:24:11 --> 00:24:17
I'm going to write that k plus
omega t.

350
00:24:20 --> 00:24:26

351
00:24:31 --> 00:24:37
i omega t plus k.

352
00:24:36 --> 00:24:42
Thank you.
i omega t plus k,

353
00:24:40 --> 00:24:46
or k plus i omega t.

354
00:24:47 --> 00:24:53
kt?
Sorry.

355
00:24:48 --> 00:24:54
So, it's k times e to the i
omega t times e to the kt.

356
00:24:57 --> 00:25:03
So, that's (k plus i omega)

357
00:25:06 --> 00:25:12
times t. Sorry.

358
00:25:10 --> 00:25:16
So, y tilda e to the kt
is k divided by,

359
00:25:17 --> 00:25:23
now I integrate this,
so it essentially reproduces

360
00:25:23 --> 00:25:29
itself, except you have to put
down on the bottom k plus i

361
00:25:30 --> 00:25:36
omega.
I'll take the final step.

362
00:25:35 --> 00:25:41
What's y tilda equals,
see, when you do it this way,

363
00:25:38 --> 00:25:44
then you don't get a messy
looking formula that you

364
00:25:42 --> 00:25:48
substitute into and that is
scary looking.

365
00:25:44 --> 00:25:50
This is never scary.
Now, I'm going to do two things

366
00:25:48 --> 00:25:54
simultaneously.
First of all,

367
00:25:49 --> 00:25:55
here, if I multiply,
after I get the answer,

368
00:25:52 --> 00:25:58
I'm going to want to multiply
it by e to the negative kt,

369
00:25:56 --> 00:26:02
right,
to solve for y tilda.

370
00:26:00 --> 00:26:06
If I multiply this by e to the
negative kt, then that just gets

371
00:26:04 --> 00:26:10
rid of the k that I put in,
and left back with e to the i

372
00:26:08 --> 00:26:14
omega t.
So, that side is easy.

373
00:26:10 --> 00:26:16
All that is left is e to the i
omega t.

374
00:26:14 --> 00:26:20
Now, what's interesting is this
thing out here,

375
00:26:18 --> 00:26:24
k plus i omega.
I'm going to take a typical

376
00:26:22 --> 00:26:28
step of scaling it.
And you scale it.

377
00:26:24 --> 00:26:30
I'm going to divide the top and
bottom by k.

378
00:26:29 --> 00:26:35
And, what does that produce?
One divided by one plus i times

379
00:26:35 --> 00:26:41
omega over k.

380
00:26:40 --> 00:26:46
What I've done is take these
two separate constants,

381
00:26:45 --> 00:26:51
and shown that the critical
thing is their ratio.

382
00:26:51 --> 00:26:57
Okay, now, what I have to do
now is take the real part.

383
00:26:57 --> 00:27:03
Now, there are two ways to do
this.

384
00:27:01 --> 00:27:07
There are two ways to do this.
Both are instructive.

385
00:27:08 --> 00:27:14
So, there are two methods.
I have a multiplication.

386
00:27:13 --> 00:27:19
The problem is,
of course, that these two

387
00:27:17 --> 00:27:23
things are multiplied together.
And, one of them is,

388
00:27:23 --> 00:27:29
essentially,
in Cartesian form,

389
00:27:26 --> 00:27:32
and the other is,
essentially,

390
00:27:29 --> 00:27:35
in polar form.
You have to make a decision.

391
00:27:35 --> 00:27:41
Either go polar,
it sounds like go postal,

392
00:27:40 --> 00:27:46
doesn't it, or worse,
like a bear,

393
00:27:45 --> 00:27:51
savage, attack it savagely,
which that's a very good,

394
00:27:52 --> 00:27:58
aggressive attitude to have
when doing a problem,

395
00:27:58 --> 00:28:04
or we can go Cartesian.
Going polar is a little faster,

396
00:28:05 --> 00:28:11
and I think it's what's done in
the nodes.

397
00:28:08 --> 00:28:14
The notes to do both of these.
They just do the first.

398
00:28:11 --> 00:28:17
On the other hand,
they give you a formula,

399
00:28:14 --> 00:28:20
which is the critical thing
that you will need to go

400
00:28:18 --> 00:28:24
Cartesian.
I hope I can do both of them if

401
00:28:21 --> 00:28:27
we sort of hurry along.
How do I go polar?

402
00:28:24 --> 00:28:30
To go polar,
what you want to do is express

403
00:28:27 --> 00:28:33
this thing in polar form.
Now, one of the things I didn't

404
00:28:32 --> 00:28:38
emphasize enough,
probably, when I talked to you

405
00:28:35 --> 00:28:41
about complex numbers last time
is, so I will remind you,

406
00:28:40 --> 00:28:46
which saves my conscience and
doesn't hurt yours,

407
00:28:43 --> 00:28:49
suppose you have alpha as a
complex number.

408
00:28:47 --> 00:28:53
See, this complex number is a
reciprocal.

409
00:28:50 --> 00:28:56
The good number is what's down
below.

410
00:28:52 --> 00:28:58
Unfortunately,
it's downstairs.

411
00:28:55 --> 00:29:01
You should know,
like you know the back of your

412
00:28:58 --> 00:29:04
hand, which nobody knows,
one over alpha.

413
00:29:03 --> 00:29:09
So that's the form.
The number I'm interested in,

414
00:29:05 --> 00:29:11
that coefficient,
it is of the form one over

415
00:29:08 --> 00:29:14
alpha.
One over alpha times alpha is

416
00:29:10 --> 00:29:16
equal to one.

417
00:29:13 --> 00:29:19
And, from that,
it follows, first of all,

418
00:29:15 --> 00:29:21
if I take absolute values,
if the absolute value of one

419
00:29:19 --> 00:29:25
over alpha times the absolute
value of this is equal to one,

420
00:29:22 --> 00:29:28
so, this is equal to one over
the absolute value of alpha.

421
00:29:26 --> 00:29:32
I think you all knew that.
I'm a little less certain you

422
00:29:29 --> 00:29:35
knew how to take care of the
angles.

423
00:29:33 --> 00:29:39
How about the argument?
Well, the argument of the

424
00:29:36 --> 00:29:42
angle, in other words,
the angle of one over alpha

425
00:29:40 --> 00:29:46
plus, because when you multiply,
angles add.

426
00:29:44 --> 00:29:50
Remember that.
Plus, the angle associated with

427
00:29:48 --> 00:29:54
alpha has to be the angle
associated with one.

428
00:29:51 --> 00:29:57
But what's that?
One is out here.

429
00:29:54 --> 00:30:00
What's the angle of one?
Zero.

430
00:29:58 --> 00:30:04

431
00:30:06 --> 00:30:12
Therefore, the argument,
the absolute value of this

432
00:30:10 --> 00:30:16
thing is want over the absolute
value.

433
00:30:14 --> 00:30:20
That's easy.
And, you should know that the

434
00:30:18 --> 00:30:24
argument of want over alpha is
equal to minus the argument of

435
00:30:23 --> 00:30:29
alpha.
So, when you take reciprocal,

436
00:30:27 --> 00:30:33
the angle turns into its
negative.

437
00:30:30 --> 00:30:36
Okay, I'm going to use that
now, because my aim is to turn

438
00:30:35 --> 00:30:41
this into polar form.
So, let's do that someplace,

439
00:30:40 --> 00:30:46
I guess here.
So, I want the polar form for

440
00:30:48 --> 00:30:54
one over one plus i times omega
over k.

441
00:31:00 --> 00:31:06
Okay, I will draw a picture.

442
00:31:04 --> 00:31:10
Here's one.
Here is omega over k.

443
00:31:09 --> 00:31:15
Let's call this angle phi.

444
00:31:12 --> 00:31:18
It's a natural thing to call
it.

445
00:31:15 --> 00:31:21
It's a right triangle,
of course.

446
00:31:18 --> 00:31:24
Okay, now, this is going to be
a complex number times e to an

447
00:31:24 --> 00:31:30
angle.
Now, what's the angle going to

448
00:31:28 --> 00:31:34
be?
Well, this is a complex number,

449
00:31:32 --> 00:31:38
the angle for the complex
number.

450
00:31:35 --> 00:31:41
So, the argument of the complex
number, one plus i times omega

451
00:31:40 --> 00:31:46
over k is how much?

452
00:31:43 --> 00:31:49
Well, there's the complex
number one plus i over one plus

453
00:31:48 --> 00:31:54
i times omega over k.

454
00:31:53 --> 00:31:59
Its angle is phi.
So, the argument of this is

455
00:31:57 --> 00:32:03
phi, and therefore,
the argument of its reciprocal

456
00:32:01 --> 00:32:07
is negative phi.
So, it's e to the minus i phi.

457
00:32:06 --> 00:32:12
And, what's A?

458
00:32:09 --> 00:32:15
A is one over the absolute
value of that complex number.

459
00:32:14 --> 00:32:20
Well, the absolute value of
this complex number is one plus

460
00:32:20 --> 00:32:26
omega over k squared.

461
00:32:24 --> 00:32:30
So, the A is going to be one
over that, the square root of

462
00:32:29 --> 00:32:35
one plus omega over k,
the quantity squared,

463
00:32:33 --> 00:32:39
times e to the minus i phi.

464
00:32:39 --> 00:32:45
See, I did that.

465
00:32:43 --> 00:32:49
That's a critical step.
You must turn that coefficient.

466
00:32:46 --> 00:32:52
If you want to go polar,
you must turn is that

467
00:32:49 --> 00:32:55
coefficient, write that
coefficient in the polar form.

468
00:32:52 --> 00:32:58
And for that,
you need these basic facts

469
00:32:54 --> 00:33:00
about, draw the complex number,
draw its angle,

470
00:32:57 --> 00:33:03
and so on and so forth.
And now, what's there for the

471
00:33:02 --> 00:33:08
solution?
Once you've done that,

472
00:33:06 --> 00:33:12
the work is over.
What's the complex solution?

473
00:33:10 --> 00:33:16
The complex solution is this.
I've just found the polar form

474
00:33:16 --> 00:33:22
for this.
So, I multiply it by e to the i

475
00:33:20 --> 00:33:26
omega t,
which means these things add.

476
00:33:25 --> 00:33:31
So, it's equal to A,
this A, times e to the i omega

477
00:33:30 --> 00:33:36
t minus i times phi.

478
00:33:37 --> 00:33:43
Or, in other words,
the coefficient is one over,

479
00:33:42 --> 00:33:48
this is a real number,
now, square root of one plus

480
00:33:47 --> 00:33:53
omega over k squared.

481
00:33:53 --> 00:33:59
And, this is e to the,
see if I get it right,

482
00:33:58 --> 00:34:04
now.
And finally,

483
00:34:00 --> 00:34:06
now, what's the answer to our
real problem?

484
00:34:05 --> 00:34:11
y1: the real answer.
I mean: the really real answer.

485
00:34:11 --> 00:34:17
What is it?
Well, this is a real number.

486
00:34:13 --> 00:34:19
So, I simply reproduce that as
the coefficient out front.

487
00:34:17 --> 00:34:23
And for the other part,
I want the real part of that.

488
00:34:20 --> 00:34:26
But you can write that down
instantly.

489
00:34:23 --> 00:34:29
So, let's recopy the
coefficient.

490
00:34:25 --> 00:34:31
And then, I want just the real
part of this.

491
00:34:28 --> 00:34:34
Well, this is e to the i times
some crazy angle.

492
00:34:32 --> 00:34:38
So, the real part is the cosine
of that crazy angle.

493
00:34:36 --> 00:34:42
So, it's the cosine of omega t
minus phi.

494
00:34:40 --> 00:34:46
And, if somebody says,

495
00:34:42 --> 00:34:48
yeah, well, okay,
I got the omega k,

496
00:34:45 --> 00:34:51
I know what that is.
That came from the problem,

497
00:34:49 --> 00:34:55
the driving frequency,
driving angular frequency.

498
00:34:52 --> 00:34:58
That was omega,
and k, I guess,

499
00:34:55 --> 00:35:01
k was the conductivity,
the thing which told you how

500
00:34:59 --> 00:35:05
quickly the heat that penetrated
the walls of the little inner

501
00:35:03 --> 00:35:09
chamber.
So, that's okay,

502
00:35:07 --> 00:35:13
but what's this phi?
Well, the best way to get phi

503
00:35:11 --> 00:35:17
is just to draw that picture,
but if you want a formula for

504
00:35:15 --> 00:35:21
phi, phi will be,
well, I guess from the picture,

505
00:35:19 --> 00:35:25
it's the arc tangent of omega,
k, divided by k,

506
00:35:23 --> 00:35:29
over one,
which I don't have to put

507
00:35:28 --> 00:35:34
in.
So, it's this number,

508
00:35:30 --> 00:35:36
phi, in reference to this
function.

509
00:35:34 --> 00:35:40
See, if the phi weren't there,
this would be cosine omega t,

510
00:35:39 --> 00:35:45
and we all know what that looks

511
00:35:44 --> 00:35:50
like.
The phi is called the phase lag

512
00:35:48 --> 00:35:54
or phase delay,
something like that,

513
00:35:51 --> 00:35:57
the phase lag of the function.
What does it represent?

514
00:35:56 --> 00:36:02
It represents,
let me draw you a picture.

515
00:36:02 --> 00:36:08
Let's draw the picture like
this.

516
00:36:05 --> 00:36:11
Here's cosine omega t.

517
00:36:08 --> 00:36:14
Now, regular cosine would look
sort of like that.

518
00:36:13 --> 00:36:19
But, I will indicate that the
angular frequency is not one by

519
00:36:18 --> 00:36:24
making my cosine squinchy up a
little too much.

520
00:36:23 --> 00:36:29
Everybody can tell that that's
the cosine on a limp axis,

521
00:36:28 --> 00:36:34
something for Salvador Dali,
okay.

522
00:36:31 --> 00:36:37
So, there's cosine of
something.

523
00:36:36 --> 00:36:42
So, what was it?
Blue?

524
00:36:37 --> 00:36:43
I don't have blue.
Yes, I have blue.

525
00:36:41 --> 00:36:47
Okay, so now you will know what
I'm talking about because this

526
00:36:46 --> 00:36:52
looks just like the screen on
your computer when you put in

527
00:36:52 --> 00:36:58
the visual for this.
Frequency: your response order

528
00:36:56 --> 00:37:02
one.
So, this is cosine omega t.

529
00:36:59 --> 00:37:05
Now, how will cosine omega t

530
00:37:04 --> 00:37:10
minus phi look?

531
00:37:07 --> 00:37:13
Well, it'll be moved over.
Let's, for example,

532
00:37:10 --> 00:37:16
suppose phi were pi over two.
Now, where's pi over two on the

533
00:37:15 --> 00:37:21
picture?
Well, what I do is cosine omega

534
00:37:18 --> 00:37:24
t minus this.
I move it over by one,

535
00:37:22 --> 00:37:28
so that this point becomes that
one, and it looks like,

536
00:37:26 --> 00:37:32
the site will look like this.
In other words,

537
00:37:30 --> 00:37:36
I shove it over by,
so this is the point where

538
00:37:33 --> 00:37:39
omega t equals pi over two.

539
00:37:39 --> 00:37:45
It's not the value of t.
It's not the value of t.

540
00:37:43 --> 00:37:49
It's the value of omega t.

541
00:37:46 --> 00:37:52
And, when I do that,
then the blue curve has been

542
00:37:50 --> 00:37:56
shoved over one quarter of its
total cycle, and that turns it,

543
00:37:55 --> 00:38:01
of course, into the sine curve,
which I hope I can draw.

544
00:38:01 --> 00:38:07
So, this goes up to there,
and then, it's got to get back

545
00:38:05 --> 00:38:11
through.
Let me stop there while I'm

546
00:38:08 --> 00:38:14
ahead.
So, this is sine omega t,

547
00:38:11 --> 00:38:17
the yellow thing,

548
00:38:13 --> 00:38:19
but that's also,
in another life,

549
00:38:16 --> 00:38:22
cosine of omega t minus pi over
two.

550
00:38:21 --> 00:38:27
The main thing is you don't
subtract, the pi over two is not

551
00:38:26 --> 00:38:32
being subtracted from the t.
It's being subtracted from the

552
00:38:32 --> 00:38:38
whole expression,
and this whole expression

553
00:38:35 --> 00:38:41
represents an angle,
which tells you where you are

554
00:38:39 --> 00:38:45
in the travel,
a long cosine to this.

555
00:38:41 --> 00:38:47
What this quantity gets to be
two pi, you're back where you

556
00:38:46 --> 00:38:52
started.
That's not the distance on the

557
00:38:49 --> 00:38:55
t axis.
It's the angle through which

558
00:38:51 --> 00:38:57
you go through.
In other words,

559
00:38:54 --> 00:39:00
does number describes where you
are on the cosine cycle.

560
00:38:58 --> 00:39:04
It doesn't tell you,
it's not aiming at telling you

561
00:39:01 --> 00:39:07
exactly where you are on the t
axis.

562
00:39:04 --> 00:39:10
The response function looks
like one over the square root of

563
00:39:09 --> 00:39:15
one plus omega over k squared
times cosine omega t minus phi.

564
00:39:13 --> 00:39:19

565
00:39:19 --> 00:39:25
And, I asked you on the problem
set, if k goes up,

566
00:39:24 --> 00:39:30
in other words,
if the conductivity rises,

567
00:39:28 --> 00:39:34
if heat can get more rapidly
from the outside to the inside,

568
00:39:34 --> 00:39:40
for example,
how does that affect the

569
00:39:38 --> 00:39:44
amplitude?
This is the amplitude,

570
00:39:42 --> 00:39:48
A, and the phase lag.
In other words,

571
00:39:47 --> 00:39:53
how does this affect the
response?

572
00:39:51 --> 00:39:57
And now, you can see.
If k goes up,

573
00:39:55 --> 00:40:01
this fraction is becoming
smaller.

574
00:39:59 --> 00:40:05
That means the denominator is
becoming smaller,

575
00:40:05 --> 00:40:11
and therefore,
the amplitude is going up.

576
00:40:12 --> 00:40:18
What's happening to the phase
lag?

577
00:40:14 --> 00:40:20
Well, the phase lag looks like
this: phi one omega over k.

578
00:40:20 --> 00:40:26
If k is going up,

579
00:40:23 --> 00:40:29
then the size of this side is
going down, and the angle is

580
00:40:28 --> 00:40:34
going down.
Now, that part is intuitive.

581
00:40:32 --> 00:40:38
I would have expected everybody
to get that.

582
00:40:36 --> 00:40:42
It's the heat gets in quickly,
more quickly,

583
00:40:40 --> 00:40:46
then the amplitude will match
more quickly.

584
00:40:44 --> 00:40:50
This will rise,
and get fairly close to one,

585
00:40:47 --> 00:40:53
in fact, and there should be
very little lag in the way the

586
00:40:53 --> 00:40:59
response follows input.
But how about the other one?

587
00:40:57 --> 00:41:03
Okay, I give you two minutes.
The other one,

588
00:41:01 --> 00:41:07
you will figure out yourself.